$\lim_{x\to\infty}\left(\frac{2^x+4^x}{4}\right)^{\frac{1}{x}}$

$y^2x^2+x^3$

$\frac{8}{x+4}=\frac{7}{x-2}$

$\frac{\sin\left(a\right)}{1-\cos\left(a\right)}\cdot\frac{\cos\left(a\right)}{\sin\left(a\right)}$

$1+6+12+8$

$2x^2-3x+7=0$

$\left(x^3-5x^2+x\:\right)\:\left(x^2\right)$